Proving that $\lim \limits_{x \to a} f(g(x)) = f \left(\lim \limits_{x \to a}g (x) \right) $ The texts that I'm using present, as a theorem, that
$$ \lim \limits_{x \to a} g (x) = b \wedge f\in\mathcal {C}^0(b) \Rightarrow \lim \limits_{x \to a} f(g(x)) = f \left(\lim \limits_{x \to a}g (x) \right) \text{;}$$
but no proof (other than "it makes intuitive sense") is offered, and I'm stumped at what approach to take to prove it. I feel I'm missing something though. Is there a straightforward proof that avoids brute force application of the $\varepsilon$-$\delta$ definition of the limit?
 A: I think it's good to see the $\varepsilon$-$\delta$ definition in action as well.
Fix $\varepsilon \gt 0$. Since $f$ is continuous at $b$, there exists some $\delta \gt 0$ such that 
$$
y \in (b - \delta, b+\delta) \quad \implies \quad f(y) \in (f(b) - \varepsilon, f(b) + \varepsilon).
$$
Now since $\lim \limits_{x \to a} \ g(x) = b$, by definition, there exists some $\gamma \gt 0$ such that 
$$
x \in (a - \gamma, a + \gamma) \smallsetminus \{ a \} \quad\implies\quad g(x) \in (b - \delta, b+\delta).
$$
Combining these two statements, we get that
$$
x \in (a - \gamma, a + \gamma) \smallsetminus \{ a \} \quad\implies\quad 
g(x) \in (b - \delta, b+\delta) \quad\implies\quad
f(g(x)) \in (f(b) - \varepsilon, f(b)+\varepsilon) .
$$
Since for every $\varepsilon \gt 0$, we showed the existence of such a $\gamma \gt 0$, we can conclude that $\lim \limits_{x \to a} \ f(g(x)) = f(b)$. 
A: Here's a formal argument:
If $\lim\limits_{x\to a}\;g(x) = b$, then for every $\epsilon\gt 0$ there exists $\delta\gt 0$ such that if $0\lt|x-a|\lt \delta$, then $|g(x)-b|\lt\epsilon$.
Since $f$ is continuous at $b$, for every $\varepsilon\gt 0$ there exists $\zeta\gt 0$ such that if $0\lt|x-b|\lt \zeta$, then $|f(x)-f(b)|\lt \varepsilon$.
In order to prove that 
$$\lim_{x\to a}\; f\bigl( g(x)\bigr) = f\left(\lim_{x\to a}g(x)\right) = f(b),\qquad\qquad{(1)}$$
we need to show that for every $\theta\gt 0$ there exists $\Delta\gt 0$ such that if $0\lt|x-a|\lt \Delta$, then $|f(g(x)) - f(b)|\lt\theta$. 
To that end, let $\theta\gt 0$. Setting $\varepsilon=\theta$, we know that there exists $\zeta\gt 0$ such that if $0\lt |x-b|\lt \zeta$, then $|f(x)-f(b)|\lt \theta$.
Setting $\epsilon=\zeta$, we know that there exists $\delta\gt 0$ such that if $0\lt |x-a|\lt\delta$, then $|g(x)-b|\lt \zeta$.
I claim that $\Delta=\delta$ will work to establish (1). Indeed, let $x$ be such that $0\lt |x-a|\lt \delta$. Then we know that $0\leq |g(x)-b|\lt \zeta$.  We have two cases: either $g(x)=b$, or $g(x)\neq b$.


*

*If $g(x) = b$, then $f(g(x))=f(b)$, so $|f(g(x))-f(b)| = 0\lt \theta$, and we are fine.

*If $g(x)\neq b$, then we know that $0\lt|g(x)-b|\lt \zeta$. Therefore, by choice of $\zeta$, we know that $|f(g(x)) - f(b)|\lt \theta$.
Thus, in either case, we conclude that $|f(g(x))-f(b)|\lt\theta$.
In summary, we have shown that if $\theta\gt 0$, then there exists $\Delta\gt 0$ such that if $0\lt|x-a|\lt \Delta$, then $|f(g(x))-f(b)|\lt\theta$. This means that
$$\lim_{x\to a}\;f\bigl(g(x)\bigr) = f(b) = f\left(\lim_{x\to a}g(x)\right),$$
as claimed. $\Box$
For an intuitive argument: the limit of $f(g(x))$ as $x\to a$ is $f(b)$ if and only if we can make $f(g(x))$ arbitrarily close to $f(b)$ for all values of $x$ near enough $a$; we know that we can ensure that if $g(x)$ is close enough to $b$, because $f$ is continuous at $b$; and we know we can make $g(x)$ always be arbitrarily close to $b$ for all values of $x$ near enough $a$, because the limit of $g(x)$ as $x\to a$ is $b$. So first, ensure that $x$ is near enough $a$ that $g(x)$ will be close enough to $b$, and this will in turn guarantee that $f(g(x))$ is as near to $f(b)$ as we required.
A: Consider a sequence $x_n$ which converges to $a$ the the sequence $g(x_n)$ converges to $b$ by $\lim_{x \to a} g(x)=b$ then $ f(g(x_n))$ converge to $f(b)$ since $f$ is continuous hence $\lim_{x\to a} f(g(x))= f(b)= f(\lim_{x \to a} g(x))$ as required.
